Question: A pot of piping hot stew has been removed from the stove and left to cool. The relationship between the elapsed time, $m$, in minutes, since the stew was removed from the stove, and the temperature of the stew, $T(m)$, measured in $^\circ C$, is modeled by the following function. $T(m)=20+50 \cdot 10^{-0.04m}$ How many minutes will it take for the stew to cool to a temperature of $30^\circ C\,$ ? Round your answer, if necessary, to the nearest hundredth.
Explanation: Thinking about the problem We want to know how many minutes, $m$, it will take for the temperature of the stew, $T(m)$, to cool to $30^\circ C\,$. So we need to find the value of $m$ for which $T(m)=30$. Substituting $30$ in for $T(m)$ in the function gives us the following equation. $30=20+50 \cdot 10^{-0.04m}$ Solving the equation We can solve the equation as shown below. $\begin{aligned}20+50 \cdot 10^{-0.04m}&=30\\\\ 50 \cdot 10^{-0.04m }&=10\\\\ 10^{-0.04m}&=0.2\\\\ -0.04m&=\log(0.2)\\\\ m&=\dfrac{\log(0.2)}{-0.04}\\\\ m&\approx 17.47\end{aligned}$ It will take $17.47$ minutes for the temperature of the stew to cool to $30^\circ C$.